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Differentiate the following polynomials.

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Note: some questions may not include all the possible terms.

$\\simplify[basic,zeroterm,zerofactor,unitfactor]{{ac[0]}x^5+{bc[0]}x^4+{cc[0]}x^3+{dc[0]}x^2+{ec[0]}x+{fc[0]}}$

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$f'(x)=$[[0]]$x^4+$[[1]]$x^3+$[[2]]$x^2+$[[3]]$x+$[[4]]

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$f''(x)=$[[5]]$x^3+$[[6]]$x^2+$[[7]]$x+$[[8]]

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$\\simplify[basic,zeroterm,zerofactor,unitfactor]{{ac[1]}x^{ap}+{bc[1]}x^{bp}+{cc[1]}x^{cp}+{dc[1]}x^{dp}+{ec[1]}x+{fc[1]}}$

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$f'(x)=$ [[0]]

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$f''(x)=$ [[1]]

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$\\simplify[basic,zeroterm,zerofactor,unitfactor]{{ac[2]}x^{ap}+{bc[2]}x^{bp}+{cc[2]}x^{cp}+{dc[2]}x^{dp}+{ec[2]}x+{fc[2]}}$

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$f'(x)=$ [[0]]

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$f''(x)=$ [[1]]

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$\\simplify[basic,zeroterm,zerofactor,unitfactor]{{ac[3]}x^{ap}+{bc[3]}x^{bp}+{cc[3]}x^{cp}+{dc[3]}x^{dp}+{ec[3]}x+{fc[3]}}$

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$f'(x)=$ [[0]]

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$f''(x)=$ [[1]]

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If $y=ax^n$,

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$\\frac{dy}{dx}=anx^{n-1}$ for all rational $n$.

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We'll take one of the terms from Part a as an example:

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$\\var{cc[0]}x^\\var{cp}$

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All we have to do to terms where $x$ is to a power of anything is times the coefficient of $x$ by the original power, and then take one away from the original power.

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If you are not familiar with this kind of work, these instructions may sound confusing, but it is much easier once you have seen it in practice.

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We take

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$\\var{cc[0]}x^\\var{cp}$

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and times $\\var{cc[0]}$ by $\\var{cp}$, to get

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$(\\var{cc[0]}*\\var{cp})x^\\var{cp}=\\simplify{{cc[0]}*{cp}x^{cp}}$.

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We then subtract one from the original power, $\\var{cp}$.

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This gives us the final answer of

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$\\simplify{{cc[0]}*{cp}x^{cp-1}}$.

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Remember, don't be confused if there is no coefficient. The fact the term is there means the coefficient must be $1$, but we don't tend to write it out as, for example $1x$, we just say $x$.

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A basic introduction to differentiation

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